Calculate the pH of a 0.0415 M aqueous solution of ethylamine (C2H5NH2, Kb = 4.3×10-4). pH...

Question

Question

Calculate the pH of a 0.0415 M aqueous solution of ethylamine (C2H5NH2, Kb = 4.3×10-4). pH =

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Answer 1

Answer #1

C2H5NH2 + H2O = C2H5NH3+ + OH-

0.0415 . 0. 0 . 0 ( initially)

0.0415-x . 0. x. x (equilibrium)

kB = ((C2H5NH3+)*(OH-))/( 0.0415-x)

4.3*10^-4 = x*x / ( 0.0415-x)

x = 4.015*10^-3 = OH-

pOH = -log OH-

= - log(4.015*10^-3)

= 2.4

pH = 14-2.4

= 11.6

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Answer 2

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